### Video Transcript

Determine the indefinite integral
of eight π to the three π₯ minus π to the two π₯ plus nine over seven π to the
power of π₯ with respect to π₯.

Now, this question might look
really nasty, and you might be starting to consider how a substitution might
help. However, itβs important to notice
that we can simplify the integrand by simply dividing each part of the numerator by
seven π to the power of π₯, remembering that we can then simply subtract the
exponents. When we do, weβre left with the
indefinite integral of eight-sevenths π to the two π₯ minus one-seventh π to the
π₯ plus nine-sevenths π to the negative π₯.

Next, we recall that we can
separate this integral. The integral of the sum of a number
of functions is equal to the sum of the integrals of each function. And we can also take any constant
factors outside of the integral and deal with them later. So, weβre looking at eight-sevenths
of the integral of π to the two π₯ with respect to π₯ minus one-seventh of the
integral of π to the π₯ with respect to π₯ plus nine-sevenths of the integral of π
to the power of negative π₯ with respect to π₯. Okay, So what next?

Well, we know that the indefinite
integral of π to the power of π₯ is π to the power of π₯ plus π. But what about the integral of π
of the power of two π₯? You might be wanting to make a
prediction as to what you think this will give. And there is a standard result that
we can quote. But letβs look at this derivation
using integration by substitution. Weβre going to let π’ be equal to
two π₯. So, dπ’ by dπ₯ is equal to two. Now, we know that dπ’ by dπ₯ is
absolutely not a fraction, but we can treat it a little like one for the purposes of
integration by substitution. And we see that a half dπ’ is equal
to dπ₯.

And then, we see that we can simply
replace two π₯ with π’ and dπ₯ with a half dπ’ and then take out this factor of
one-half. And all we need to do now is
integrate π to the power of π’ with respect to π’. Well, thatβs simply a half times π
to the power of π’. But of course, since π’ is equal to
two π₯, we can say that the indefinite integral of π to the power of two π₯ is a
half times π to the power of two π₯ plus a constant of integration. Letβs call that π΄. And this is great because it
provides us with the general result for the integral of π the power of ππ₯ for
real constants π. Itβs one over π times π to the
power of ππ₯.

We can use this result. And we see that the integral of π
to the power of negative π₯ is one over negative one times π to the power of
negative π₯ plus π. So, popping this altogether, we see
that our indefinite integral is eight-sevenths times a half π to the two π₯ plus π΄
minus a seventh times π to the π₯ plus π΅ plus nine-sevenths times one over
negative one times π to the power of negative π₯ plus πΆ. Distributing our parentheses and
combining all our constants, and weβre left with the solution as being four-sevenths
π to the power of two π₯ minus π to power of π₯ over seven minus nine-sevenths π
to the power of negative π₯ plus π·.